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Triss [41]
2 years ago
10

Calculate the resistance of a piece of aluminum wire with a diameter of 100 mils and a length of two miles, at 68°F. Hint: Be su

re to first convert mils to cmils and use the K value for aluminum found in the reference. (Round the FINAL answer to two decimal places.)
Mathematics
1 answer:
Oduvanchick [21]2 years ago
4 0
First you need to know the formula of the Resistance(R) of a conductor, which is:

R =  \frac{K*L}{CM} -- (A)

Where,
R = Resistance of the wire = ?
K = Specific resistance of the wire(In this case Aluminum at 68°F) = 17.02 ohm-circular-mil-foot
L = Length of the wire(in feet)
CM = Circular mil Area(In inches)


Now in order to use the above formula(A), we have to first convert TWO quantities:
1) The length of the wire(from miles to feet)
2) Mils to Circular mil Area

1.
The length of the wire given = 2 miles
The length of the wire(In feet) = 10560 ft
2.
Since the mils given = 100 mils.
Therefore,
Circular mils(Circular mil Area) = mils x mils = 100 x 100 = 10000 

Now plug-in the values in equation(A):

A=> R = \frac{17.02 * 10560}{10000 }

R = 17.97 ohms 

Ans: Resistance of the aluminum wire = 17.97 ohms.

-i
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A manufacturing company produces valves in various sizes and shapes. One particular valve plate is supposed to have a tensile st
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Answer:

z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413    

p_v =2*P(z>1.413)=0.158  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

Step-by-step explanation:

Data given and notation  

\bar X=5.0611 represent the sample mean

\sigma=0.2803 represent the population standard deviation for the sample  

n=42 sample size  

\mu_o =5 represent the value that we want to test

\alpha=0.1 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 5, the system of hypothesis would be:  

Null hypothesis:\mu = 5  

Alternative hypothesis:\mu \neq 5  

If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413    

P-value

Since is a two sided test the p value would be:  

p_v =2*P(z>1.413)=0.158  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

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The future worth (F) of the current investment (P) that has an interest (i) that is compounded annually is calculated through,
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where n is the number of compounding period. Substituting the given values,
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Answer:

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Step-by-step explanation:

15*3=45 slices of pie in total,

each pie is cut into 9 slices,

45/9=5 pies

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